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Added some universal algebra general nonsense
Andrew Stacey
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Obligatory n-lab reference: adjoint functor theorems

Figuring out when functors had adjoints or not was something I did a lot of in Comparative Smootheology (section 8).

Edit: Thought I'd expand on my comment to Andrew Critch's answer. A simple application of the Special Adjoint Functor Theorem is to universal algebra where it becomes:

Theorem Let $D$ be a category that has finite products, is co-complete, is an $(E, M)$ category where $E$ is closed under finite products, is $E$-co-well-powered, and its finite products commute with filtered co-limits. Let $V$ be a variety of algebras. Let $F$ be a category with co-equalisers. Let $G : F \to DV$ (here, $DV$ is the category of $V$-algebra objects in $D$) be a covariant functor. Then the following statements are equivalent.

  1. $G$ has a left adjoint.
  2. The composition $|G| : F \to D$ of $G$ with the forgetful functor $DV \to D$ has a left adjoint.

In particular, if we take $D$ to be $Set$, the category of sets, then we obtain the following (which can be found in any text book on universal algebra), in which the variety of algebras $V$ is identified with its category of models in $Set$:

Corollary Let $F$ be a co-complete category, $V$ a variety of algebras. For a covariant functor $G : F \to V$, the following statements are equivalent.

  1. $G$ has a left adjoint.
  2. $G$ is representable by a co-$V$-algebra object in $F$.
  3. $|G|$ is representable by an object in $F$.

(And, of course, all of this can be turned round for adjoint pairs of contravariant functors)

In further particular, if $G : F \to V$ preserves underlying sets then $|G|$ is representable (by the initial $F$-object) and so $G$ has a left adjoint.

Andrew Stacey
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